question_answer

The area of the plane region bounded by the curves \[x+2{{y}^{2}}=0\]and \[x+3{{y}^{2}}=1\]is equal to

A) \[1/3\]                           

B) \[2/3\]    

C) \[4/3\]               

D)        \[5/3\]

Correct Answer: C

Solution :

Parabola: \[{{y}^{2}}=\frac{-x}{2}\]and \[{{y}^{2}}=\frac{1}{3}(1-x)\] On solving, we get \[x=-2,\,y=\pm 1\] \[\therefore \] required Area             \[=2\left[ \frac{1}{\sqrt{3}}\int\limits_{-2}^{1}{\sqrt{(1-x)}\,dx-\frac{1}{\sqrt{2}}\int\limits_{-2}^{0}{\sqrt{-x}dx}} \right]\] \[=2\left\{ \left[ \frac{1}{\sqrt{3}}\times \frac{-2}{3}{{(1-x)}^{3/2}} \right]_{-2}^{1}-\left[ \frac{1}{\sqrt{2}}\times \frac{-2}{3}{{(-x)}^{3/2}} \right]_{-2}^{0} \right\}\]\[=2\left\{ \left( \frac{2}{3\sqrt{3}}.3\sqrt{3} \right)-\left( \frac{2}{3\sqrt{2}}.2\sqrt{2} \right) \right\}=\frac{4}{3}.\]